Tonka trucks and springs...?

Question

A child places a 2.2 kg Tonka truck against a horizontal spring (k = 2300 N/m). Assume that the truck is on a frictionless, horizontal surface. In an attempt to launch the truck, the child pushes it back so that the spring is compressed 15 cm, then releases the truck.

a. How much energy is stored in the spring just before the truck is released?

b. What is the velocity of the truck after it is "launched" (leaves the spring)?

c. If the spring inadvertently becomes hooked to the bumper of the truck, what will be the period of the truck's oscillation?

d. As in part c, if the spring inadvertently becomes hooked to the bumper of the truck, calculate the location of the truck 0.31 seconds after the truck is released.

Bonus question! What's oscillation? :P

Thanks a ton!

Answer 1

♠ Energy stored is E=0.5*k*A^2, where k=2300 N/m, A=15cm = 0.15 m; thus

(a) E=0.5*2300*0.15^2 =25.875 J;  

♣ since energy conserves we get E=0.5*m*V^2, where m=2.2 kg, hence speed

(b) V=√(2E/m)= √(2*25.875/2.2) =4.850 m/s;

(c)♦ period T=2pi*√(m/k) = 2*3.1416*√(2.2/2300) =0.1943 s;  

♥ the equation of motion of our toy is: x=A*cos(2pi*t/T) =A*cos(t*√(k/m));

(d) x= 15*cos(0.31*√(2300/2.2)) =-12.392 cm;

Bonus: oscillation is a periodic motion about equilibrium position with amplitude A, where A is max deflection from equilibrium position x=0;

Synonyms: rocking, reeling, swaying, swinging, anything to-n-fro;

Btw x=-12.392 cm means that the toy has passed equilibrium and is positioned at 12.392 cm on the other side stretching the spring;